General Solution of Vector ODE in Given Form The ODE in question, given below, is straightforward to solve:
$$\mathbf{\ddot r} = -\omega^2 \mathbf r$$
for $\mathbf r \in \Bbb R^2$.
However, I seek to show that the general solution may be written as,
$$\mathbf r = \mathbf a \sin (\omega t + \alpha) + \mathbf b \cos (\omega t + \alpha)$$
where $\mathbf a$ and $\mathbf b$ are constant and are orthogonal to each other (and $\alpha$ is a real constant).
 A: We know that the solution can be given as 
$$
\mathbf r = \mathbf A \sin (\omega t ) + \mathbf B \cos (\omega t)
$$
where $\mathbf A$, $\mathbf B$ have no special relationship. Introducing the phase angle $α$ and using trigonometric identities for $\sin((ωt+α)-α)$ etc. one gets
$$
\mathbf r=\mathbf a\sin(ωt+α)+\mathbf b\cos(ωt+α)
$$
with
\begin{align}
\mathbf a &= \mathbf B \sin(α) + \mathbf A \cos(α) \\
\mathbf b &= \mathbf B \cos(α) - \mathbf A \sin(α) 
\end{align}
and their scalar product is 
\begin{align}
\langle \mathbf a,\, \mathbf b\rangle 
&= (\|\mathbf B\|^2-\|\mathbf A\|^2)\sin(α)\cos(α)+⟨\mathbf A,\mathbf B⟩(\cos^2(α)-\sin^2(α))
\\
&= \frac12(\|\mathbf B\|^2-\|\mathbf A\|^2)\sin(2α)+⟨\mathbf A,\mathbf B⟩\cos(2α)
\end{align}
Thus one can find the angle $α$ with the desired properties as half the angle of the point 
$$
(\cos(2α),\sin(2α))=\Bigl(\frac12(\|\mathbf A\|^2-\|\mathbf B\|^2),\,⟨\mathbf A,\mathbf B⟩\Bigr)
$$ 
(relative to origin and positive horizontal axis in a Cartesian plane).
